prove that sin0-cos0+1/sin0+cos0-1=1/sec0-tan0
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Step-by-step explanation:
sinO-cosO+1/sinO+cosO-1
devide by cosO in each term
tanO+secO-1/tanO-secO+1
={(tanO+secO)-1}(tanO-secO)/(tanO-secO+1)(tanO-secO)
=tan²O-sec²O-tanO+secO/(tanO-secO+1)(tanO-secO)
=-1-tanO+secO/(tan-secO+1)(tanO-secO)
=-1/tanO-secO
=1/secO-tanO
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